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Let $R$ be a transitive relation. Is the symmetric closure of $R$ also transitive?

user107952
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  • Go through the axioms. – J.G. Dec 23 '20 at 23:06
  • $R = {(a, b), (b,c)}.$ Symmetric closure means $R^s = {(a, b), (b, a),(b, c), (c, b}$. R is transitive., $R^s$ is not. To be transitive, $R^s$ must include $(a, a), (b, b), (a, c)$ et al. – amWhy Dec 23 '20 at 23:07
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    @amWhy In your example $R$ is not transitive, as it must also include $(a,c)$ to be transitive. – aefrrs Dec 24 '20 at 02:16

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Counterexample, adapted from @amWhy's attempt based on an observation by @aefrrs: if $a\ne b\ne c\ne a$, the transitive relation $\{(a,\,b),\,(b,\,c),\,(a,\,c)\}$ has symmetric closure $\{(a,\,b),\,(b,\,c),\,(a,\,c),\,(b,\,a),\,(c,\,b),\,(c,\,a)\}$, which isn't transitive as it doesn't own $\{a,\,a\}$. As @NoahSchweber notes, a simpler example, whose transitivity is vacuous, is $\{(a,\,b)\}$ with $a\ne b$.

J.G.
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